Sciencemadness Discussion Board

Copper Sulfate from Magnesium Sulfate

rollercoaster158 - 21-4-2012 at 14:42

Well, I decided to take a stab at making copper sulfate from epsom salt (magnesium sulfate). My setup was a small glass container with about 120ml of water and 2-3 ounces of epsom salt (This stuff dissolves impressively!). I stirred the solution until it turned clear. I then coiled some copper wire and put it in the bottom of the solution, as the anode. My cathode was a ring of aluminum foil at the top, in the solution. I powered it up with 6 volts DC, and stirred it frequently.

MgSO4 + H2O + Cu = CuSO4 + MgOH + H

The solution turned blue, but a cloudy blue and not a clear blue we know copper sulfate solutions to contain. It got so thick that I couldn't see the copper coil. I turned it off, removed the electrodes, and let it sit. The blue cloud settled to the bottom, and the top was clear. I poured off as much of the top as I could, then left the bottom to evaporate. A few days later it was dry, so I broke the crystal out and smashed it into a coarse powder. The problem is that this powder was only a sky blue, was completely opaque, had white spots in it, clear crystals of magnesium hydroxide and some bits of copper. When I added the powder to water, it broke into a finer powder, but wouldn't dissolve or produce wisps of blue in the water. Even with stirring, and vigorous shaking, nothing would make it dissolve. Does anyone know why this happened, and if my "copper sulfate" is either a hybrid salt, or just a failure?

ManBearSwine - 21-4-2012 at 15:24

Copper Hydroxide is much less soluble than Magnesium Hydroxide. The reaction that takes place is:
2 H2O + Cu --> H2 to Cu(OH)2
To my knowledge, you cannot product copper sulfate without the use of either sulfuric acid or another soluble copper salt.
If you washed the copper hydroxide with water to remove the impurities then neutralized it with sulfuric acid, you would have a solution of copper sulfate, which you could then crystallize by adding it to alcohol or acetone.

DerAlte - 21-4-2012 at 16:24

First balance your equations and get your valencies right:
MgSO4 + H2O + Cu == > CuSO4 + MgOH + H
MgSO4 + 2H2O + Cu ==> CuSO4 + Mg(OH)2 + H2

But you cannot write an electrolysis as a single reaction. There are usually at least two reactions – anode and cathode, and possibly also a reaction in the bulk solution. The electrolysis of a dilute solution of a neutral salt is usually considered as merely electrolysis of water. (Concentrated solutions are more complicated). Water ionizes as 2H2O < ==> OH- + H3O+

There are also Mg++ and SO4— ions at both anode and cathode.

Of those ions next to the anode the hydroxide is attracted to and its electron transferred to the anode; and so is the SO4— ion.. discharging two electrons. This results in two very reactive radicals:

2OH- ==> H2O + O* + 2e-
SO4-- == > SO4* + 2e-

The SO4* radical would immediately react with water to produce sulphuric acid:
SO* + H2O == > 2H+ SO4-- + 2O*

The net effect is thus the generation of oxygen. If is does not react with the anode, then two radials combine to produce oxygen gas: O* + O* -- > O2 – as with a Pt anode.
But copper is far more easily oxidized than platinum; it is attacked:
Cu(s) + O* ==> CuO

And this immediately reacts with the acid produced to give copper sulphate near the anode, the copper being oxidized to Cu++ and going into solution. Effectively, then copper sulphate builds up a concentration near the anode.

At the cathode a similar process happens with the cations Mg++ and H3O-. However, Al does not produce a hydride under these circumstances. Essentially, the cathode reaction is just H3O+ + e- ==> H2O + H* then 2H* == > H2. This leaves OH- ions next to the cathode – ie the solution becomes alkaline. But there are also Mg++ ions nearby, so Mg(OH)2, being insoluble, precipitates as a white powder.

So far the reactions are as indicated in your equation (corrected). But in time the Cu++ ions also migrate to the cathode. The solubility of Cu(OH)2 is less than that of Mg(OH)2:
The solubility products are:
Copper(II) hydroxide Cu(OH)2 2.6 x 10-19
Magnesium hydroxide Mg(OH)2 1.8 x 10-11

The net result is that copper hydroxide is produced at the cathode in preference to the magnesium. The only way to avoid this is to use a cell with a diaphragm. See the thread

http://www.sciencemadness.org/talk/viewthread.php?tid=12941&...

Regards
Der Alte

rollercoaster158 - 21-4-2012 at 17:30

Thanks for the help. I am relatively new to electrolysis, so this should help alot. Another thing though: I tried a similar experiment on Ammonium Dihydrogen Phosphate from a crystal growing kit. Chemical formula NH4H2PO4, here is the Wikipedia article: http://en.wikipedia.org/wiki/Ammonium_dihydrogen_phosphate
Same setup with a copper wire anode and aluminum foil ring cathode, I electrolyzed a solution of it for about 30 mins. It produced a cloudy blue precipitate similar to the one I described earlier, but much faster. I was hoping to get the decomposition products of ammonia and phosphoric acid, but this was not the case. The aluminum was plated with copper (It looked like copper, but was an ugly brown. Might be CuO) and there was quite a mess as copper flaked off of the wire. I turned off power, and the cloudy solution settled to the bottom. Is this copper phosphate from reaction with the liberated phosphoric acid, or more copper hydroxide?

Pyridinium - 21-4-2012 at 21:12

Quote: Originally posted by rollercoaster158  
It produced a cloudy blue precipitate similar to the one I described earlier, but much faster. I was hoping to get the decomposition products of ammonia and phosphoric acid, but this was not the case. The aluminum was plated with copper (It looked like copper, but was an ugly brown. Might be CuO) and there was quite a mess as copper flaked off of the wire. I turned off power, and the cloudy solution settled to the bottom. Is this copper phosphate from reaction with the liberated phosphoric acid, or more copper hydroxide?


It would help to know how much NH4H2PO4 you dissolved.

At the anode you have Cu++ ions going into solution. Before they even get near the cathode let's consider what happens to the H2PO4- ion in solution

H2PO4- (aq) <-------------> H+ (aq) + HPO4-- (aq)
HPO4-- (aq) <-------------> H+ (aq) + PO4--- (aq)

The pKa for these is increasingly large (small ionization), but copper could react at either stage to form a precipitate. I'm guessing the more likely one is CuHPO4 because there would be much, much more HPO4-- in solution than PO4---.

There was a brief bit about CuHPO4 at this link
http://journals.iucr.org/e/issues/2009/12/00/fi2089/fi2089bd...


There also forms the CuH2PO4+ ion in solution.

Cu++ (aq) + H2PO4- (aq) <------> CuH2PO4+ (aq)

According to some formation constants I found, this should be a lot more abundant than CuHPO4 (s), but I don't know if the CuH2PO4+ ion would form any actual stable precipitates with anything. I don't know how much it dissociates or anything else. Maybe someone can dig up an old reference to this.

If you have copper hydroxide, it should yield black copper oxide on standing, but this can depend on the conditions under which it was precipitated.

At the cathode you would have copper plating out, but sometimes it doesn't look like copper because of the high surface area of the metal particles. There are some reactions involving Cu2O and Cu(OH)2, but the end result of these at the cathode is the plating out of copper metal. You could get formation of these compounds due to OH- at the cathode (as already noted by DerAlte). Perhaps the material plating out on the cathode is actually a mixture of Cu metal and the oxide, which could also explain its appearance.

If any of this doesn't make sense or if I missed anything someone can correct it- it's getting late and I'm getting tired.

[Edited on 22-4-2012 by Pyridinium]

rollercoaster158 - 22-4-2012 at 06:21

Well, I know it did produce something. I dissolved about 60 grams of monoammonium phosphate in about 150ml of water, and had to stir for 10 minutes or so to fully dissolve it. The electrolysis started off uneventful, but it had a light blue color after a while. Before I knew it, the tiny particles in the solution clumped into a thick, sky blue goop at the top of my container in less than 10 seconds. I ran it longer and got my container half full with this strange stuff. I took out the electrodes and the aluminum looked really ugly, and the copper had dissolved quite a bit. I am debating whether I should pour off the top, or if it has useful things in it too. Guess I'll let the whole thing evaporate and see how that goes.

AJKOER - 27-4-2012 at 05:26

A non-electrolysis approach:

1. Prepare dilute H2SO4 using Oxalic acid from the Epsom Salt. Separate out the Magnesium Oxalate.

MgSO4.7H2O + H2C2O4 --> MgC2O4 (s) + H2SO4 + 7 H2O

2. Add copper oxide to the dilute acid. Depending on how dilute, the longer the reaction may take.

CuO + H2SO4 --> CuSO4 + H2O

To quote Wikipedia: "Copper sulfate is produced industrially by treating copper metal with hot concentrated sulfuric acid or its oxides with dilute sulfuric acid. For laboratory use, copper sulfate is usually purchased." Please note this last comment.

A simple process and basically pure CuSO4, but more expensive than just buying it.
-----------------------------------------

A possible faster variation of this synthesis.

1. Dissolve Cu or CuO in Oxalic acid:

Cu + H2C2O4 --> CuC2O4 (s) + H2 (g)

Source: "A dictionary of chemistry and mineralogy: with an account of the ...", by Arthur Aikin, Charles Rochemont Aikin, Vol 2, page 194, to quote: "Oxalate of Copper. This acid dissolves copper both metallic and oxidated. The salt is scarcely soluble in water. Most of the other solutions of this metal are decomposed by the oxalic acid, and an insoluble oxalate falls down". http://books.google.com/books?pg=PA194&lpg=PA194&dq=...

2. Add Na2C2O4 to dissolve the CuC2O4 precipitate.

Source: per Watt's Vol 4, page 257 (in the library): "Neutral cupric Oxalate, ....(according to Lowe, Jaliresb. 1880, p. 243), is a light greenish-blue precipitate, insoluble in water, nearly or quite insoluble, in oxalic acid, but easily soluble in the neutral oxalates of ammonium, potassium and sodium").

3. Add MgSO4 and separate out the insoluble MgC2O4.

CuC2O4 + MgSO4 --> CuSO4 + MgC2O4 (s)


[Edited on 27-4-2012 by AJKOER]